+1 vote
in Machine Kinematics by (38.0k points)
A small connecting rod of mass 1.5 kg is suspended in a horizontal plane by two wires 1.25 m long. The wires are attached to the rod at points 120 mm on either side of the centre of gravity. If the rod makes 20 oscillations in 40 seconds, find the radius of gyration of the rod about a vertical axis through the centre of gravity.

(a) 107 mm

(b) 207 mm

(c) 307 mm

(d) 407 mm

I had been asked this question in unit test.

Asked question is from Closely-coiled Helical Spring and Compound Pendulum topic in section Simple Harmonic Motion of Machine Kinematics

1 Answer

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by (106k points)
The correct choice is (a) 107 mm

For explanation: Given : m = 1.5 kg ; l = 1.25 m ; x = y = 120 mm = 0.12 m

Since the rod makes 20 oscillations in 40 s, therefore frequency of oscillation,

n = 20/40 = 0.5 Hz

Let kG = Radius of gyration of the connecting rod.

We know that frequency of oscillation (n),

0.5 = 1/2πkG √gxy/l = 1/2πkG √9.81 x 0.12 x 0.12/1.25 = 0.0535/k

kG = 0.0535/0.5 = 0.107 m = 107 mm.

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